Proving that the derivative always diverges faster than the original function

This is actually wrong. For this function: there are values of x arbitrarily close to $0$ such that $\frac{f'(x)}{f(x)} = 0$. If you want an explicit expression, take $f(x) = \frac1x + sin\Big(\frac1x\Big)$

The most you can show is that $f'/f$ is unbounded in any neighborhood of $a$.

With $a < x < y$, it follows by the mean value theorem that there exists a point $\xi_{x,y}$ between $x$ and $y$ such that

$$\log f(x) - \log f(y) = \frac{f'(\xi_{x,y})}{f(\xi_{x,y})}(x - y),$$

and

$$\lim_{x \to a+}\frac{f'(\xi_{x,y})}{f(\xi_{x,y})} = \lim_{x \to a+} \frac{\log f(x) - \log f(y)}{x-y} = \infty.$$

Hence, on any interval $(a,y]$ no matter how small we can find a sequence of points $(\xi_n)$ such that $f'(\xi_n)/f(\xi_n) \to \infty.$ The mean value theorem is non-constructive with respect to the intermediate point, so we cannot determine that $\xi_n \to a$ or more generally that $\lim_{x \to a} \xi_{x,y}= a.$ This shows, at least, that $f'/f$ must be unbounded in any neighborhood of $x=a.$

The answer by @Eulerr is an example where $f'/f$ is unbounded but the limit does not exist.